# Conjecture about the powers of 2 in the prime factorizations of the series $a_n = n(n+1)(n+2)$

I found the following law and would like to know what do you think about it and if anyone can explain why this is so. Also, is this already known and proven?

Consider the following series: $$1\times2\times3, 2\times3\times4, 3\times4\times5,\ldots,n(n+1)(n+2)$$

or more compactly:

$$a_n = n(n+1)(n+2)$$

Lets take the prime factorization for every element of $a_n$:

$$2\times3, 2^3\times3, 2^2\times3\times5,2^3\times3\times5,\ldots$$

Lets take all the powers of $2$ in every prime factorization in the above series and create a sequence out of them, and then group the elements of that sequence in tuples with size $4$. The result of this we will call $p_1$. $$p_1 = (1,3,2,3),(1,4,3,4),(1,3,2,3),(1,5,4,5),\ldots$$

It seems that every $4$-tuple with odd index in $p_1$ is $(1,3,2,3)$.

Lets remove all $4$-tuples with odd positions from $p_1$ and call the result $p_2$:

$$p_2 = (1,4,3,4),(1,5,4,5),(1,4,3,4),(1,6,5,6),\ldots$$

Now $p_2$ will have $(1,4,3,4)$ at every odd position. If we take the $p_3$ sequence to have only those elements of $p_2$ that are with even indices, $p_3$ will have $(1,5,4,5)$ at every odd position. And so on... $p_n$ will have $(1, n+2, n+1,n+2)$ at every odd position.

• Probably shouldn't use the word partition here, because that has a very specific meaning. Maybe just "group the elements in sets of four." – Thomas Andrews Nov 29 '14 at 0:24
• Thanks for the advice. I fixed this. – mzdravkov Nov 29 '14 at 0:32
• Are you familiar with congruences? Discuss separately the cases $n\equiv k \mod 8$ for $k=1,\ldots, 8$ and everything will fall into place. – user138530 Nov 29 '14 at 0:59
• Speaking of $a_n=n(n+1)(n+2)$ I notice that most often Pell´s equation with $d=a_n+1$ has a much higher (lowest) solution than $d=a_n$ when $a_n+1$ is a prime, Pell´s equation being $x^2-dy^2=1$. Peter Košinár calculated for me that there are three exceptions in the first 800 primes. (“Search for a counterexample for a Pell’s equation conjecture”). – Mikael Jensen Apr 5 '15 at 14:43

Let $n=8k+1$. Then $a_{8k+1}=(8k+1)(8k+2)(8k+3)=2(8k+1)(4k+1)(8k+3)$, which shows that $2\mid a_{8k+1}$ but $4\not\mid a_{8k+1}$. Similarly,
$$a_{8k+2}=8(4k+1)(8k+3)(2k+1)$$ $$a_{8k+3}=4(8k+3)(2k+1)(8k+5)$$ $$a_{8k+4}=8(2k+1)(8k+5)(4k+3)$$
show that indeed $(1,3,2,3)$ always repeat itself.
The next part is showing that $a_{16k+5}$, $a_{16k+6}$, $a_{16k+7}$ and $a_{16k+8}$ have the same property. Indeed $$a_{16k+5}=2(16k+5)(8k+3)(16k+7)$$ $$a_{16k+6}=16(8k+3)(16k+7)(2k+1)$$ $$a_{16k+7}=8(16k+7)(2k+1)(16k+9)$$ $$a_{16k+8}=16(2k+1)(16k+9)(8k+5),$$ so the pattern $(1,4,3,4)$ will also repeat forever.
In general, \begin{align} a_{8km+4k-3}&=(8km+4k-3)(8km+4k-2)(8km+4k-1)\\&=2(8km+4k-3)(4km+2k-1)(8km+4k-1)\\ \\ a_{8km+4k-2}&=(8km+4k-2)(8km+4k-1)(8km+4k)\\ &=8k(4km+2k-1)(8km+4k-1)(2m+1)\\ \\ a_{8km+4k-1}&=(8km+4k-1)(8km+4k)(8km+4k+1)\\ &=4k(8km+4k-1)(2m+1)(8km+4k+1)\\ \\ a_{8km+4k}&=(8km+4k)(8km+4k+1)(8km+4k+2)\\&=8k(2m+1)(8km+4k+1)(4km+2k+1) \end{align} generates $(1, 3+v(k), 2+ v(k), 3+v(k))$, denoting by $v(k)$ the power of $2$ in the factorization of $k$.
• Note you can greatly reduce your general result by noting that $a_{4n-3},a_{4n-2},a_{4n-1},a_{4n}$ yields $(1,3+v(n),2+v(n),3+v(n))$. This is because $n=k(2m+1)$ in the above, and thus $v(n)=v(k)$. – Thomas Andrews Nov 29 '14 at 1:43